Problem: $ F = \left[\begin{array}{rrr}-1 & 4 & 4 \\ 0 & -2 & -2\end{array}\right]$ $ D = \left[\begin{array}{rr}5 & -1 \\ 3 & 4 \\ 5 & 4\end{array}\right]$ What is $ F D$ ?
Explanation: Because $ F$ has dimensions $(2\times3)$ and $ D$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ F D = \left[\begin{array}{rrr}{-1} & {4} & {4} \\ {0} & {-2} & {-2}\end{array}\right] \left[\begin{array}{rr}{5} & \color{#DF0030}{-1} \\ {3} & \color{#DF0030}{4} \\ {5} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{5}+{4}\cdot{3}+{4}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{5}+{4}\cdot{3}+{4}\cdot{5} & ? \\ {0}\cdot{5}+{-2}\cdot{3}+{-2}\cdot{5} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{5}+{4}\cdot{3}+{4}\cdot{5} & {-1}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{4}+{4}\cdot\color{#DF0030}{4} \\ {0}\cdot{5}+{-2}\cdot{3}+{-2}\cdot{5} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{5}+{4}\cdot{3}+{4}\cdot{5} & {-1}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{4}+{4}\cdot\color{#DF0030}{4} \\ {0}\cdot{5}+{-2}\cdot{3}+{-2}\cdot{5} & {0}\cdot\color{#DF0030}{-1}+{-2}\cdot\color{#DF0030}{4}+{-2}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}27 & 33 \\ -16 & -16\end{array}\right] $